Question

Tính S :     \(\frac{1}{2}\)+ \(\frac{1}{4}\)+ \(\frac{1}{8}\)+ ... + \(\frac{1}{1024}\).

Answer 1

S=\(\frac{1}{2}\)+\(\frac{1}{4}\)+\(\frac{1}{8}\)+.............+\(\frac{1}{1024}\)

S=1-1/2+1/2-1/4+1/4-1/8+.........+1/512-1/1024

S=1-1/1024

S=1023/1024

Vậy s=1023/1024

Answer 2

Ta có : \(\frac{1}{2}=1-\frac{1}{2};\frac{1}{4}=\frac{1}{2}-\frac{1}{4};\frac{1}{8}=\frac{1}{4}-\frac{1}{8};...;\frac{1}{1024}=\frac{1}{512}-\frac{1}{1024}\)

Vậy \(S=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{1024}\)

           = \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{8}+...+\frac{1}{512}-\frac{1}{1024}\)

           = \(1-\frac{1}{1024}=\frac{1023}{1024}\)